13-day Yunnan Travel Notes (Plus Shanghai Diet Diary & Ale¡¯s Hand-drawn Travel Diary)

13-day Yunnan Travel Notes (Plus Shanghai Diet Diary & Ale¡¯s Hand-drawn Travel Diary)

Matematika Nilai dari lim x →∞0 4x² - 2x+1 2x² - 4x+4​

Nilai dari lim x →∞0 4x² - 2x+1 2x² - 4x+4​

[tex]\large\text{$\begin{aligned}\lim_{x\to\infty}\:\frac{4x^2-2x+1}{2x^2-4x+4}=\boxed{\ \bf2\ }\end{aligned}$}[/tex]

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Pendahuluan

Limit Fungsi Rasional Sukubanyak

Nilai limit dari fungsi rasional sukubanyak pada saat [tex]x[/tex] mendekati tak hingga, yang dinyatakan oleh:

[tex]\begin{aligned}\lim_{x\to\infty}\frac{a_0x^m+a_1x^{m-1}+{\dots}+a_{m-1}x+a_m}{b_0x^n+b_1x^{n-1}+{\dots}+b_{n-1}x+b_n}\end{aligned}[/tex]

adalah:

[tex]\begin{aligned}\begin{cases}\infty\,,&\textsf{jika }m > n\\\\\dfrac{a_0}{b_0}\,,&\textsf{jika }m = n\\\\0\,,&\textsf{jika }m < n\end{cases}\end{aligned}[/tex]

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Pembahasan

Kita akan menentukan nilai limit dari:

[tex]\begin{aligned}\lim_{x\to\infty}\:\frac{4x^2-2x+1}{2x^2-4x+4}\end{aligned}[/tex]

Secara singkat, kita bisa tentukan nilai limit tersebut, yaitu:

[tex]\begin{aligned}\lim_{x\to\infty}\:\frac{4x^2-2x+1}{2x^2-4x+4}=\boxed{\ \bf2\ }\end{aligned}[/tex]

karena [tex]m=n=2[/tex], dan [tex]a_0/b_0=4/2=\bf2[/tex].

Jika ingin dijabarkan lebih lanjut:

[tex]\begin{aligned}&\lim_{x\to\infty}\:\frac{4x^2-2x+1}{2x^2-4x+4}\\{=\ }&\lim_{x\to\infty}\left(\frac{4x^2-2x+1}{2x^2-4x+4}\times\frac{1/x^2}{1/x^2}\right)\\{=\ }&\lim_{x\to\infty}\frac{4-\dfrac{2}{x}+\dfrac{1}{x^2}}{2-\dfrac{4}{x}+\dfrac{4}{x^2}}\\{=\ }&\frac{4-0+0}{2-0+0}\\{=\ }&\boxed{\ \bf2\ }\end{aligned}[/tex]

Atau menggunakan aturan L’Hopital, karena memenuhi bentuk pecahan tak tentu, yaitu ∞/∞.

[tex]\begin{aligned}&\lim_{x\to\infty}\:\frac{4x^2-2x+1}{2x^2-4x+4}\\{=\ }&\lim_{x\to\infty}\:\frac{\frac{d}{dx}\left(4x^2-2x+1\right)}{\frac{d}{dx}\left(2x^2-4x+4\right)}\\{=\ }&\lim_{x\to\infty}\:\frac{8x-2}{4x-4}\\&...\text{ masih memenuhi bentuk }\tfrac{\infty}{\infty}\\&...\text{ terapkan aturan L'H\^{o}pital}\\{=\ }&\lim_{x\to\infty}\:\frac{\frac{d}{dx}(8x-2)}{\frac{d}{dx}(4x-4)}\\{=\ }&\lim_{x\to\infty}\:\frac{8}{4}\ =\ \lim_{x\to\infty}\:2\\{=\ }&\boxed{\ \bf2\ }\end{aligned}[/tex]

[tex]\blacksquare[/tex]

KESIMPULAN

[tex]\large\text{$\begin{aligned}\therefore\ \lim_{x\to\infty}\:\frac{4x^2-2x+1}{2x^2-4x+4}=\boxed{\ \bf2\ }\end{aligned}$}[/tex]

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